Physics Help Forum Michelson-Morley I measuere new facts

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 Sep 28th 2012, 05:05 AM #11 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 The jpeg is not very clear, so I don't understand. It seems that you have two pipes now, one with 10 times as much energy as the other, and hence the KE of the masses from the second pipe is greater than the first: V_00 > V_0. That seems obvious, though it has nothing to do with the speed of the earth relative to some point A. But perhaps I am not interpreting your drawing correctly. I suggest that you take the time to carefully write out your explanation - it would make it much easier for me or others to follow your reasoning and offer feedback. One specific piece of advice: in the movies you mention several times that lightning goes to ground because that's the easiest path, and present this as if it has something to do with your theory. But you never tie it together - you say "energy wants to go where it's easiest," but give no explanation as to what that has to do with your theory. Last edited by ChipB; Sep 28th 2012 at 06:27 AM.
Sep 28th 2012, 11:03 AM   #12
Banned

Join Date: Sep 2012
Posts: 64
Dear ChipB Thank good dialog below I explain ALL

Please check my pdf ( thank You for nice and good dialog I very respect this )

we schould start speek what I can do with my knwledge

I want don't want stop Einstein AGE but for me nothing is relative (einstein has problemnot me )

see what point A mean for classic I need start research many new facts
in this area ( I hope not alone ) on my home page I show very good patents I'm good engineer

Attached Files
 pointA.pdf (530.3 KB, 1 views)

Last edited by tesla2; Sep 28th 2012 at 11:09 AM.

 Sep 28th 2012, 11:54 AM #13 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 I don't believe the 6 equations you wrote are sufficient to determine v0. Try this: suppose I give you the following data: v1= 1 m/s v2 = 10m/s m= 1 Kg Delta E1L = 0.5 N-m Delta E1R = 0.5 N-m Delta E2L = 50 N-m Delta E2R = 50 N-m What do you conclude is the value for v0? I believe it's indeterminate.
 Sep 28th 2012, 01:33 PM #14 Banned   Join Date: Sep 2012 Posts: 64 v1= 1 m/s v2 = 10m/s ( see what I write below this is not true ) m= 1 Kg Delta E1L = 0.5 N-m Delta E1R = 0.5 N-m Delta E2L = 50 N-m Delta E2R = 50 N-m m =1kg ok E = 1 N-m energy in pipe 1 V1 = 1m/s ( ok ) Delta E1L = 0.5 N-m ( ok ) Delta E1R = 0.5 N-m (ok ) 10 E !!! = 10 N-m energy in pipe 2 V2 = square root from 10 ( not 10m/s !!!!!) Delta E2L = 50 N-m ??? 5 N-m !!! Delta E2R = 50 N-m ??? 5 N-m !!! My equations are ok You make small mistake in volume I understand Your poitn of view or not ? ( Delta E1L = Delta E1R condition ) for my equations mean that rocket has zero speed relation to Point A
 Sep 28th 2012, 02:02 PM #15 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 I thought making E2 = 100 E1 would make the math easier to follow. Hence I have v2 = 10 V1. In any event, the point is that there is no way from this data to determine V0. V0 could be 0, or 100m/s, or -10^6 m/s; there's no way to tell.
 Sep 30th 2012, 08:28 AM #16 Banned   Join Date: Sep 2012 Posts: 64 Ok good poit of view I start use math my equations need need only sign modif. all You can see page 4 pdf I add this page after Your Post exist on more condition that not fit to my equations ( situation III ) www.maroszvsnewton.cba.pl/pointA1_1.pdf Thank You for help by Your opinion
 Oct 1st 2012, 08:41 AM #17 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 I'm good with your situation 1 and situation 2, but I do not understand situation 3. The velocities of "after explosion" as observed by A should be v0-v1 and v0-v2, just as in suitiation 1 (although at relativistic speeds the velocities are not additive). But this is moot, as your assertion was that somehow the observer on the rocket ship could determine his v0, and your PDF doesn't indicate how that person could do this. To the observer on the rocket ship the velocities are always v1 and v2. Please address how the observer on the rocket could measure v0 without observing A itself. Attached Thumbnails
 Oct 18th 2012, 11:05 PM #18 Banned   Join Date: Sep 2012 Posts: 64 I finish my experiment with camera below data Dear Cheap Thank You for Your Hard disscusion with me I made final experiment at the end of experiment I translate why my camera see not the same power of light this what I measure in my experiment PLEASE SEE YELLOW ARROW MEAN ( VELOCITY EARTH AROUND SUN 30KM/S ) ( copy and open in new window ) www.maroszvsnewton.cba.pl/whatexplain.pdf how to build own experiment in home material list : www.maroszvsnewton.cba.pl/materiallist.pdf main idea how to prepare : www.maroszvsnewton.cba.pl/instrukcja.pdf what mean average brightness ( explain photoshop tool ) : www.maroszvsnewton.cba.pl/photoshop10.pdf final camera set : www.maroszvsnewton.cba.pl/cameraset.pdf Real Experiment 24h test (6h ,6h,6h,6h ) see how I measure Earth velocity and Earth around own axis velocity ( sign change my camera feel this change) START Poland 8 of October 18:00 18:00 Poland 50 degree Wide please see where was Earth and Poland position West /East camera position I mark Blue/Green Line in my pdf ( please see how camera see Earth own rotation - Earth around own Axis give in Poland 1,8km/s line velocity ) 24h earth rotation = equator distance Poland has lower distance I count 1,8 km/s ( prepedicular direction to Earth around Sun velocity 18:00 think about Velocity Vectore in point where I measured ) 18:00 ( 08 October Start ) www.maroszvsnewton.cba.pl/eksa.pdf 24:00 (6h later ) www.maroszvsnewton.cba.pl/eksb.pdf below movie translate how photoshop 10 see this what I measured my camera :I cooperate with Mr Butch from USA ( movie show 5 picture west and 5 picture East ) near 0.4 light point different after finish above I made 23-24 degree test ( before I start I counted angle using Autocad) 23-24 mean that my camera measure on Red line direction see drawing in below pdf www.maroszvsnewton.cba.pl/eksc.pdf 6:00 ( 6h later ) after I make 5 picure West and 5 East I just make decission to set camera ( automatic program - please take picture every 4 minutes ) 4 minutes = time that Earth change position 1 degree ( so below You have data - West East picture and special East light measure every 4 minutes) www.maroszvsnewton.cba.pl/eksd.pdf 12:00 finsh www.maroszvsnewton.cba.pl/eksF.pdf Theory explain - Why my camera see not the same power of light in West or East ? part 1 : part2 : Last edited by tesla2; Oct 18th 2012 at 11:10 PM.

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